Verified answer

First thing you have to do is know the derivative of 7^(sin(x))

7^(sin(x)) = e^(ln(7^(sin(x))))

= e^(sin(x) * ln(7))

d/dx 7^(sin(x)) = ln(7) * cos(x) * e^(sin(x) * ln(7))

= ln(7) * cos(x) * 7^(sin(x))

Hence, using L'Hospital's Rule, the limit is equal to:

lim as x -> 0 of ln(7) * cos(x) * 7^(sin(x)) / 4 = ln(7) / 4

If the equation is

7^(sinx) - 1/(4x)

That's the same as

lim(x->0) 7^(sinx) - lim(x->0) 1/(4x)

In which case, you only need to use L'Hospital's Rule on the second equation, as the first simplifies to 1. However, you should notice that 1/x approaches positive or negative infinity as x becomes 0, depending on which way you approach it from. In this case, there is no solution.

If the problem is

7^(sinx) - x/4

Then just substitute 0 in to get 1. In either case, L'Hospital's Rule is not needed.

do you not know that the derivative of a^(b(x)) is [a^b] b '(x) ln a...{ ln 7 / 4}

i don't speak math!